Let $h$ be a differentiable function with $h(-2)=-7$ and $h'(-2)=-8$. What is the value of the approximation of $h(-1.9)$ using the function's local linear approximation at $x=-2$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-7.9$ (Choice B) B $-7.8$ (Choice C) C $-7.7$ (Choice D) D $-7.6$
Answer: The local linear approximation of $h$ at $x=-2$ is achieved using the equation of the line tangent to $h$ at $x=-2$. Let $L(x)$ represent this equation. We can find $L(x)$ using the general formula for the tangent to the graph of function $u$ at $x=a$ : $y=u'(a)(x-a)+u(a)$ [Is there a way to find this formula without memorizing?] In our case, $L(x)=h'(-2)(x+2)+h(-2)$. Plugging $h(-2)=-7$ and $h'(-2)=-8$, we obtain $L(x)=-8(x+2)-7$. To approximate $h(-1.9)$, all we need is to plug $x=-1.9$ into $L(x)$. $\begin{aligned} L(-1.9)&=-8(-1.9+2)-7 \\\\ &=-8(0.1)-7 \\\\ &=-7.8 \end{aligned}$ In conclusion, the approximation of $h(-1.9)$ using the function's local linear approximation at $x=-2$ is $-7.8$.